To determine which solution has the highest freezing point among the given options, we need to understand the concept of colligative properties, particularly the depression of freezing point. The freezing point depression depends on the number of solute particles in a solution, not on their chemical nature.

The depression of freezing point can be expressed by the equation:

$$\Delta T_f = K_f \cdot m \cdot i$$

where:

• $$\Delta T_f$$ is the freezing point depression,
• $$K_f$$ is the cryoscopic constant (freezing point depression constant),
• $$m$$ is the molality of the solution, and
• $$i$$ is the van't Hoff factor, the number of particles the solute splits into in solution.

Since all solutions are equimolal, the important factor here is the van't Hoff factor (i). A lower van't Hoff factor will result in a smaller decrease in freezing point, meaning the freezing point will be relatively higher.

Let's analyze each option:

• Option A: C₆H₅NH₃Cl (aniline hydrochloride) dissociates into two ions: $$C_6H_5NH_3^+$$ and $$Cl^-$$.

  $$i = 2$$

• Option B: Ca(NO₃)₂ dissociates into three ions: $$Ca^{2+}$$ and two $$NO_3^-$$ ions.

  $$i = 3$$

• Option C: La(NO₃)₃ dissociates into four ions: $$La^{3+}$$ and three $$NO_3^-$$ ions.

  $$i = 4$$

• Option D: C₆H₁₂O₆ (glucose) does not dissociate into ions in an aqueous solution.

  $$i = 1$$

Among the given options, glucose (Option D) has the lowest van't Hoff factor, $$i = 1$$. Therefore, the least number of particles results in the least depression of the freezing point. Hence, an aqueous solution of glucose will have the highest freezing point among the options provided.